5.6 liter of helium gas at STP is adiabatically compressed to 0.7 liter. Taking initial temperature

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11.Thermodynamics

hard

$5.6$ liter of helium gas at $STP$ is adiabatically compressed to $0.7$ liter. Taking the initial temperature to be $T _1$, the work done in the process is

A

$\frac{9}{8} R T_1$

B

$\frac{3}{2} R T_1$

C

$\frac{15}{8} R T_1$

D

$\frac{9}{2} R T_1$

(IIT-2011)

Solution

$V_1=5.6 l, T_1=273 K, P_1=1 a t m$

$\gamma=\frac{5}{3}(F \text { or monoa } \rightarrow \text { micgas })$

The number of moles of gas is $n=\frac{5.6 l}{22.4 l}=\frac{1}{4}$

Finally (after adiabatic compression)

$V_2=0.7 l$

For adiabatic compression $T_1 V_1^{\gamma-1}=T_2 V_2^{\gamma-1}$

$\therefore T_2=T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1}=T_1\left(\frac{5.6}{0.7}\right)^{\frac{5}{3}-1}=T_1(8)^{2 / 3}$

$=4 T_1$

We know that work done in adiabatic process is

$W=\frac{n R \Delta T}{\gamma-1}=\frac{9}{8} R T_1$

Standard 11

Physics

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